Hi there,
I'd just like to check if my logic for rationalising this denominator containing a radical is correct. I have the following to solve:
\(\displaystyle \large{\frac{x}{\sqrt[4]{x}\,}}\)
Which I have then solved as follows (and this is listed as the correct answer in my book):
\(\displaystyle \large{\frac{x}{\sqrt[4]{x}}\,\cdot\, \frac{\sqrt[4]{x^3}}{\sqrt[4]{x^3}}\,=\frac{x\sqrt[4]{x^3}}{\sqrt[4]{x^4}}\,=\frac{x\sqrt[4]{x^3}}{x}=\sqrt[4]{x^3}}\,\)
I'd just like to check that my answer of \(\displaystyle \large{{\sqrt[4]{x^3}}\,}\) is correct as the x's in the numerator and denominator have effectively cancelled themselves out, so you are therefore left with \(\displaystyle \large{{\sqrt[4]{x^3}}\,}\) or in its full form... \(\displaystyle \large{\frac{x\sqrt[4]{x^3}}{x}\,}\)
I'd just like to check if my logic for rationalising this denominator containing a radical is correct. I have the following to solve:
\(\displaystyle \large{\frac{x}{\sqrt[4]{x}\,}}\)
Which I have then solved as follows (and this is listed as the correct answer in my book):
\(\displaystyle \large{\frac{x}{\sqrt[4]{x}}\,\cdot\, \frac{\sqrt[4]{x^3}}{\sqrt[4]{x^3}}\,=\frac{x\sqrt[4]{x^3}}{\sqrt[4]{x^4}}\,=\frac{x\sqrt[4]{x^3}}{x}=\sqrt[4]{x^3}}\,\)
I'd just like to check that my answer of \(\displaystyle \large{{\sqrt[4]{x^3}}\,}\) is correct as the x's in the numerator and denominator have effectively cancelled themselves out, so you are therefore left with \(\displaystyle \large{{\sqrt[4]{x^3}}\,}\) or in its full form... \(\displaystyle \large{\frac{x\sqrt[4]{x^3}}{x}\,}\)
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